Question

A body of mass m makes an elastic collision with another identical body at rest.Show that if the collision is not head-on,the bodies go at right angle to each other after the collision.

Answer 1

Since it is not a head on collision,the two bodies move in different dimensions.

Let $v_1,v_2\to$ velocities of the bodies vector collision.

Since,the collision is elastic,applying law of conservation of momentum on X=direction

$m{u}_1+m\times 0=m{v}_{1}cos\alpha +m{v}_{1}cos\beta$

$\Rightarrow v_{1}cos\alpha +v_{2}cos\beta =u_1...\left(i\right)$

Putting law of conservation of momentum in y direction $O=m{v}_{1}sin\alpha v_{2}sin\beta ...\left(ii\right)$

Again $\left(\frac{1}{2}\right)m{u}_1^2+0=\left(\frac{1}{2}\right)m{v}_1^2+\left(\frac{1}{2}\right)m\times v_2^2$

$\Rightarrow u_1^2=v_1^2+v_2^2...\left(iii\right)$

Squaring equation (i)

$v_1^2+v_2^2=v_1^{2}co{s}^2\alpha +v_2^{2}co{s}^2\beta +2v_1{v}_{2}cos\alpha cos\beta$

$\Rightarrow v_1^{2}si{n}^2\alpha +v_2^{2}si{n}^2\beta +2v_1{v}_{2}cos\alpha cos\beta$

$\Rightarrow 2v_1^{2}si{n}^2\alpha =2\times v_1\times \frac{v_{1}sin\alpha }{sin\beta }cos\alpha cos\beta$

$\Rightarrow sin\alpha sin\beta =cos\alpha cos\beta$

$\Rightarrow cos\alpha .cos\beta -sin\alpha cos\beta$

$\Rightarrow cos(\alpha +\beta )=0=cos{90}^{\circ }$

$\Rightarrow \alpha +\beta ={90}^{\circ }$